1. Limit_Problem Ex_RP

$\displaystyle lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$

2. Note that:

$\displaystyle \left(\frac{n+1}{2n}\right) \left(\frac{2n+1}{n+1}\right)$

$\displaystyle =\frac{2n+1}{2n}=1+\frac{1}{2n}$

$\displaystyle \lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

Notice the similarity to the famous 'e' limit?.

$\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}=e$

So, what is the limit of $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

3. Hint :
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

4. Originally Posted by Raoh
Hint :
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

$\displaystyle e^{\frac{1}{2}}$

$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$

5. Originally Posted by Raoh
Hint :
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=?$

$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

6. Originally Posted by Raoh
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

therefore your limit must be $\displaystyle e^{\frac{1}{2}}$.

$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$

8. Originally Posted by gilyos
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
i believe
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$

9. $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n = 0$
Why?

Intuition:

$\displaystyle (\frac{1}{2} + \frac{1}{2n})$
is always less than 1, and then raise it to any power greater than one, and it will get smaller.

10. Originally Posted by gilyos
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
Given the discussion in this thread, a bit of thought might have suggested writing $\displaystyle \left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n$ from which the limit is obvious.